Question

A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle.

(Hint: remember Corollary 1--the area of an equilateral triangle is 1/4 s2 √3.)

Answer 1

Answer:

$\frac{3}{2} \pi -\frac{9}{4} \sqrt{3}$

Step-by-step explanation:

So we know that a hexagon is made up of 6 equilateral triangles so if one side of the hexagon is equal to 3 than all the sides of one of the triangles that makes up that hexagon is equal to 3. We also know that the angles of an equilateral triangle all equal 60 degrees. So from this information we can find the area of a segment formed by a side of this hexagon inscribed in a circle.

First:

In order to find the area of a segment you need to subtract the area of the sector by the area of the triangle so we will need to find the area of a sector. The area of this sector can be found using this formula:

$\frac{60}{360} \pi (3)^2$ this the equals: $\frac{3}{2} \pi$

Second:

Now you will need to find the area of the triangle. In the hint given it tells us how to find the area. All you need to do is substitute the variables for numbers. In this case in the formula $\frac{1}{4} s^2\sqrt{3}$ s stands for side. When you substitute s for 3 the equation will look like this:

$(\frac{1}{4}) 3^2\sqrt{3}$ this equals: $\frac{9}{4} \sqrt{3}$

Third:

Now all you need to do is subtract. This leaves you with the answer: $\frac{3}{2} \pi -\frac{9}{4} \sqrt{3}$

Here's a picture of my scratch paper in case it's needed.